# Furnishing a contradiction in my proof involving Lagrange's Theorem

## The Attempt at a Solution

I'm trying to show that Case 1 implies that $\tau \in H$, and since $\tau$ was an arbitrarily chosen 3-cycle, then H must contain all 3-cycles, thus contradicting that H has 6 elements.

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Deveno
first of all, what is σ? it appears to be an arbitrary 3-cycle, but that isn't clear.

but going on that assumption, note that one of the elements of H is e.

therefore σ is in Hσ. similarly σσ is in Hσσ.

but since Hσ = Hσσ, σσ = τkσ, so σ = σσσσ = τkσσσ = τk.

thus σ is in H, which means <σ> = {σ,σσ,e} is in H.

but σ is an arbitrary 3-cycle, so....